$25\,mL\,\frac{M}{5}\,NaOH\,$ will neutralise $25\,mL\,\frac{M}{5}\,HCl$

$75 - 25 = 50\,mL\,\frac{M}{5}\,HCl$ will remain.

Total volume will be $75 + 25 = 100\,mL$

$50\,mL\,\frac{M}{5}\,HCl$ is diluted to $100\,mL$

$[{H^ + }] = [HCl] = \frac{M}{5} \times \frac{{50}}{{100}} = \frac{M}{{10}}$

$pH =  - {\log _{10}}[{H^ + }] =  - {\log _{10}}\frac{M}{{10}} = 1$